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What are the roots of $y^3 - ky^2 - k=0$ for an arbitrary constant $k$?

I haven't been able to find a root, so synthetic division couldn't be used. I have tried factorization as well.

Hendrix
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devxcholera
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    Question was closed as duplicate while I was answering, so I will leave it here: I don't know if this solves the problem, but:

    $$(y-a)(y-b)(y-c)=y^3-(a+b+c)y^2+(ab+ac+bc)y-abc$$

    Now, if $a,b$ anc $c$ are the roots of your polynomial p(y), we must have $p(y)=(y-a)(y-b)(y-c)$, therefore:

    $$a+b+c=-k$$ $$ab+ac+bc=0$$ $$abc=-k$$

    Can this system be solved?

     – David Jul 30 '19 at 16:14
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    By the way, this question is not a duplicate, as it asks for a very particular case – David Jul 30 '19 at 16:14
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    @David: yes, elimination will lead to the nice equation $z^3-kz^2-k=0$. –  Jul 30 '19 at 16:24
  • @YvesDaoust Oops! I knew that's what you get when the coefficients of the polynomial are different. I expected some simplification to be possible when we have that beautiful $k$ there – David Jul 30 '19 at 16:29

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Hint

If you apply the method for the cubic equation, you will first find $$\Delta=-k^2 \left(4 k^2+27\right) \quad < 0 \qquad \forall k$$ So, there is only one real root.

Continuing the described method, we have $p=-\frac{k^2}{3} < 0$ and $q=-\frac{2}{27} k^3-k$ and continue with the hyperbolic solution.

Claude Leibovici
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